Question: Let $f(x, y, z) = zx^2 + y^3$ and $g(t) = (t, \sin(3t), \sin(2t))$. $h(t) = f(g(t))$ $h'(t) = $
Solution: Formula The multivariable chain rule says that $\dfrac{dh}{dt} = \nabla f(g(t)) \cdot g'(t)$. The $g'(t)$ part is how much a change in $t$ will cause the input to $f$ to move, and the $\nabla f(g(t))$ part is how much $f$ will change in response to this update to its input. [What's the intuition behind the formula?] Applying the formula We want to find $h'(t) = \nabla f(g(t)) \cdot g'(t)$. $\begin{aligned} &g(t) = (t, \sin(3t), \sin(2t)) \\ \\ &g'(t) = (1, 3\cos(3t), 2\cos(2t)) \\ \\ &\nabla f = (2xz, 3y^2, x^2) \\ \\ &\nabla f(g(t)) = \left( 2t\sin(2t), 3\sin^2(3t), t^2 \right) \end{aligned}$ Substituting: $\begin{aligned} h'(t) &= 2t\sin(2t) + 9\cos(3t)\sin^2(3t) + 2t^2\cos(2t) \end{aligned}$ Answer $h'(t) = 2t\sin(2t) + 9\cos(3t)\sin^2(3t) + 2t^2\cos(2t)$